Question

An object with a mass M = 250 g is at rest on a plane that makes an angle θ = 30 o above the horizontal. The coefficient of kinetic friction between M and the plane is μk = 0.100. Mass M is attached by a string to another mass, m = 200 g, which hangs freely. When mass m has fallen 30.0 cm, its speed is:

(A) 83 cm/s
(B) 59 cm/s
(C) 160 cm/s
(D) 72 cm/s
(E) 48 cm/s

Answer 1

Answer:

v = 79.2 m/s

Solution:

As per the question:

Mass of the object, m = 250 g = 0.250 kg

Angle, $\theta = 30^{\circ}$

Coefficient of kinetic friction, $\mu_{k} = 0.100$

Mass attached to the string, m = 0.200 kg

Distance, d = 30 cm = 0.03 m

Now,

The tension in the string is given by:

$Mgsin\theta + \mu_{k}Mgcos\theta + Ma = T$        (1)

Also

T = m(g + a)

Thus eqn (1) can be written as:

$Mgsin\theta + \mu_{k}Mgcos\theta + Ma = m(g - a)$

$Mgsin\theta + \mu_{k}Mgcos\theta + Ma = mg + ma$

$mg - Mgsin\theta - \mu_{k}Mgcos\theta = (M - m)a$

$a = \frac{0.2\times 9.8 - 0.250\times 9.8\times sin30^{\circ} - 0.1\times 0.250\times 9.8\times cos30^{\circ}}{0.250 - 0.200}$

$a = 10.45\ m/s^{2}$

Now, the speed is given by the third eqn of motion with initial velocity being zero:

$v^{2} = u^{2} + 2ad$

where

u = initial velocity = 0

Thus

$v = \sqrt{2ad}$

$v = \sqrt{2\times 10.45\times 0.03} = 0.792\ m/s$